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Let $\mathcal{F}\subseteq\mathcal{P}(X)$ an algebra. Then for all sequence $\{E_k\}\subseteq\mathcal{F}$ exists a disjoint sequence $\{F_k\}\subseteq\mathcal{F}$ such that

1) $F_k\subseteq E_k$ for all $k\in \mathbb{N}$

2) $\bigcup_{k=1}^{+\infty} E_k=\bigcup_{k=1}^{+\infty} F_k$.

It is sufficient to consider the sequence \begin{align} F_1& =E_1 \\ F_n & =E_n\setminus \bigcup_{k=1}^{n-1} E_k. \end{align} Now, we observe that \begin{equation} \bigcup_{k=1}^{n} E_k=\bigcup_{k=1}^{n} F_k. \end{equation} At this point how can I conclude 2) ?

Thanks!

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I don't think the equality of the finite unions is actually a useful step along the way.

Instead, prove $\bigcup E_k\subseteq\bigcup F_k$ and $\bigcup F_k\subseteq\bigcup E_k$ separately by considering an arbitrary element of the left side and showing it must be on the right side too.

There's no need to make it an indirect proof.

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Hint: Suppose not. Let $n$ be minimal such that $$ \bigcup_{k =1}^{n} E_k \neq \bigcup_{k=1}^{n} F_k. $$ Since $F_k \subseteq E_k$ for all $k$ there is thus some $x \in E_n \setminus F_n$ (by the minimality of $n$). But $F_n = \ldots$ which leads to the desired contradiction.

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    $\begingroup$ This seems to be missing a step arguing that if the infinite unions are unequal, then there will also be some of the finite unions that are unequal. (And arguing that does not appear to be much easier than proving the entire goal directly). $\endgroup$ – Henning Makholm May 4 '18 at 11:16

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